Fast Pre-Equilibria, Intermediates and Catalysis

A mechanistic elementary law may legitimately contain an intermediate. The aim here is to predict an experimentally useful overall law in terms of reactant, product or catalyst concentrations, so relate that intermediate to those species before reporting the final expression. Some intermediates are detectable and measurable; the reason for substitution is not that they have no concentration or can never be observed.

For fast reversible 2 NO ⇌ N₂O₂ followed by slow N₂O₂ + O₂ → 2 NO₂, the slow step is bimolecular: rate = k₂[N₂O₂][O₂]. Omitting [O₂] contradicts its specified elementary character. The first equilibrium gives k₁[NO]² ≈ k₋₁[N₂O₂], so rate ≈ (k₂k₁/k₋₁)[NO]²[O₂]. Define kobs = k₂k₁/k₋₁ only after showing the substitution.

If step 1 is fast and reversible and step 2 is slow, step 1 stays effectively at equilibrium throughout. Setting the forward rate equal to the reverse rate gives an expression for the intermediate in terms of ordinary reactants, which is then substituted into the slow step.

The approximation requires equilibration in the first step to be rapid compared with removal of its intermediate by the later step. For the NO dimer example, k₂[O₂] must be small compared with k₋₁ so most dimers can re-equilibrate instead of being immediately consumed. “Fast” alone is not a universal proof. If removal is substantial, the intermediate need not maintain its equilibrium concentration and a steady-state or fuller kinetic treatment is needed.

A negative order arises when a product of the fast step appears in the denominator of the intermediate expression, so adding that product suppresses the rate. Fractional orders often signal an equilibrium involving dissociation, such as a diatomic splitting into atoms. Neither result is an error; both are mechanistic evidence.

For a fast X₂ ⇌ 2 X equilibrium, K = [X]²/[X₂], giving [X] = √(K[X₂]). A slow X + Y → products then predicts rate = k₂√K [X₂]^1/2[Y]. The half-order originates in solving the equilibrium expression, not in a collision involving half a molecule.

For fast A ⇌ B + C and slow B + D → E, [B] = K[A]/[C], so rate = k₂K[A][D]/[C]. Doubling C halves the rate under this approximation because C suppresses the supply of B. The inverse law cannot be extrapolated to C = 0, where it would incorrectly predict infinite rate; the approximation breaks down.

A homogeneous catalyst is in the same phase as the reactants and appears explicitly in the steps, consumed early and regenerated later, which is why it can appear in the rate law. A heterogeneous catalyst works on a surface through adsorption, reaction and desorption, and saturation of that surface is the usual cause of observed zero-order behaviour.

A heterogeneous catalyst's exposed active sites can become occupied. At low reactant pressure, greater coverage often raises rate; near saturation, additional reactant changes coverage little, giving an approximately zero-order regime. Poisoning instead removes or blocks useful sites and lowers activity. These are different mechanisms, even though both involve a surface. Catalysts can change selectivity as well as speed by providing a particular path, while leaving equilibrium thermodynamics for the same net reaction unchanged.